Java 如何通过url调用CXF webservice时传递参数
我创建了一个CXF webservice示例:Java 如何通过url调用CXF webservice时传递参数,java,web-services,cxf,Java,Web Services,Cxf,我创建了一个CXF webservice示例: @WebService public interface InterfaceWebService { boolean doLogin(@WebParam(name="username")String username,@WebParam(name="password")String password); } 服务器代码: public class WebServer { protected WebServer() throw
@WebService
public interface InterfaceWebService {
boolean doLogin(@WebParam(name="username")String username,@WebParam(name="password")String password);
}
服务器代码:
public class WebServer {
protected WebServer() throws Exception {
// START SNIPPET: publish
System.out.println("Starting Server");
WebServiceImpl implementor = new WebServiceImpl();
String address = "http://192.168.0.76:9000/sample";
Endpoint.publish(address, implementor);
// END SNIPPET: publish
}
public static void main(String args[]) throws Exception {
new WebServer();
System.out.println("Server ready...");
Thread.sleep(5 * 60 * 5000);
System.out.println("Server exiting");
System.exit(0);
}
}
WebServiceImpl类
@WebService(endpointInterface = "com.nextenders.services.InterfaceWebService",
serviceName = "sample")
public class WebServiceImpl implements InterfaceWebService{
@Override
public boolean doLogin(String username, String password) {
//Here some business logic call
return true;
}
}
现在我尝试通过以下url调用此Web服务:
http://192.168.0.76:9000/sample/services/doLogin?username=abc&password=abc
但是我得到了wsdl-xml结构。但我只需要特定的方法结果!!。
我做错什么了吗?如何在CXF webservice中传递参数 问题解决了。我忘了输入服务名并尝试访问direct方法 http://192.168.0.76:9000/sample/services/login_service/doLogin?username=abc&password=abc